tag:blogger.com,1999:blog-3776665703590564248.post1189297626222228054..comments2020-03-24T23:04:24.364-04:00Comments on Return To Excellence: SAT Practice QuizReturnToExcellence.nethttp://www.blogger.com/profile/18149859814999976879noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-3776665703590564248.post-83655152360950520342016-06-15T22:54:27.943-04:002016-06-15T22:54:27.943-04:00All correct answers although the arithmetic of #1 ...All correct answers although the arithmetic of #1 should be 9 x 1 = 9 not 9 x 11 = 99. See my response to PB above - 19,20,21 would work also as would any other set that went up by 10 (e.g., 29,30,31) although I used 9,10,11 also.RTEnoreply@blogger.comtag:blogger.com,1999:blog-3776665703590564248.post-90150619352052755692016-06-15T22:50:27.304-04:002016-06-15T22:50:27.304-04:00Answers to quiz:
1. (A) zero j, k n = 9 10...Answers to quiz:<br /><br /><br />1. (A) zero j, k n = 9 10, 11. 9 x 11 = 99. Units digit of 10 is 0.<br /><br /><br />2. (C) 10 10 + X = 15; X = 5; 2X = 10<br /><br /><br />3. (E) Grammatically correct<br /><br /><br />4. (A) Grammatically correct<br />JORnoreply@blogger.comtag:blogger.com,1999:blog-3776665703590564248.post-9021222287528984562015-09-19T19:50:39.925-04:002015-09-19T19:50:39.925-04:00Congratulations - excellent work.
All I can add t...Congratulations - excellent work.<br /><br />All I can add to your solution is that 19,20,21 would work also as would any other set that went up by 10 (e.g., 29,30,31) although I used 9,10,11 also.<br />RTEnoreply@blogger.comtag:blogger.com,1999:blog-3776665703590564248.post-71570457414448039542015-09-19T19:40:28.585-04:002015-09-19T19:40:28.585-04:00In regards to the math problem #1, I didn't do...In regards to the math problem #1, I didn't do anything special just trial and error. I think the problem there, is that the description of the problem was somewhat confusing and it took me 5-6 minutes of reading and rereading to understand what they were asking. 9, 10, 11 were three integers that worked.<br /><br /> Problem #2<br /><br /> 10 + X = 10 + 5<br /><br />X = 10 + 5 - 10<br /><br />X = 5<br /><br />2x = 2x5 = 10<br />PBnoreply@blogger.comtag:blogger.com,1999:blog-3776665703590564248.post-68390348718355683072015-09-17T19:29:08.888-04:002015-09-17T19:29:08.888-04:00From instructions - (please include your reasoning...From instructions - (please include your reasoning or explain how you got your answers) – @ least for the math ones. For all would be even better.<br />RTEnoreply@blogger.comtag:blogger.com,1999:blog-3776665703590564248.post-53130611055418794182015-09-17T19:26:38.104-04:002015-09-17T19:26:38.104-04:00Doug, here are my answers:
1. a
2. c
3. d
4. c
Doug, here are my answers:<br /><br />1. a<br />2. c<br />3. d<br />4. c<br />PBnoreply@blogger.comtag:blogger.com,1999:blog-3776665703590564248.post-70572434925630133442015-09-17T19:25:14.060-04:002015-09-17T19:25:14.060-04:00Doug - OK, you found some toughies.
1) I can...Doug - OK, you found some toughies. <br /><br />1) I can't figure this one out (likely due to my lack of math in school - Algebra 1 & plane Geometry. I took logic in college which was easy). Anyway, from what I recall, consecutive integers are whole numbers in sequence, such as 1, 2, 3, 4. The 'product' means multiplication. The only way to get a product of 9 would be 1 X 9 or 3 X 3 (or -1 X -9 or -3 X -3). Since all are marked < rather than a mix of < and >, I can't figure it out. Now if the integers were squares, it would work. If j is 12, k is 22 and n is 32, it would work. 1 squared is 1. Three squared is 9 and 1 X 9 = 9. But that is not an option.<br />2) C - 10<br />3) E<br />4) A<br />SC Businessmannoreply@blogger.com